Completeness of an exponential family The question is this: Does there exist an integrable function $f\colon\mathbb R\to\mathbb R$ such that $f$ differs from $0$ on a set of nonzero Lebesgue measure and 
\begin{equation}
 \int_{\mathbb R}g_\theta(x)f(x)\,dx=0 \tag{1}
\end{equation}
for all real $\theta$, where $g_\theta(x):=e^{-(\theta x-1)^2/2}$? 
More generally, one may ask this: Does there exist a nonzero tempered distribution $F$ on $\mathbb R$ such that $\langle g_\theta,F\rangle=0$ for all real $\theta$? The latter question can of course be restated as follows: Does there exist a nonzero tempered distribution $\hat F$ on $\mathbb R$ such that $\langle h_\tau,\hat F\rangle=0$ for all real $\tau$, where $h_\tau(t):=e^{i\tau t-\tau^2 t^2/2}$? 
One may note that $(1)$ will hold for all $\theta>0$ if $f=I_{(0,\infty)}-c$, where $I_{(0,\infty)}$ is the indicator function of the set $(0,\infty)$ and $c=\frac1{\sqrt{2\pi}}\int_0^\infty e^{-(u-1)^2/2}\,du$. However, the question is whether $(1)$ can hold for (an essentially nonzero $f$ and) all real $\theta$.  
 A: Here is an idea of a partial answer; cf. the answer by Christian Remling. Assume that $(1)$ holds for all $\theta$ in (say) a right neighborhood of $0$ (including $0$), where $|f(x)|=O(1/(1+|x|^k))$ for each $k>0$ (note that the latter condition is not satisfied by the function $f=I_{(0,\infty)}-c$ in the example in the question). Then one can differentiate the integral in $(1)$ in $\theta$ to the right of $0$ to get 
\begin{equation}
 \left. \int_{\mathbb R}\frac{\partial^k g_\theta(x)}{\partial\theta^k} \right|_{\theta=0}\,f(x)\,dx=0,  \tag{2}
\end{equation}
for all $k=0,1,\dots$. 
Expanding $g_\theta(x)/g_0(x)=e^{(\theta x)-(\theta x)^2/2}$ into the Maclaurin series, we have 
\begin{equation*}
 \left. \frac{\partial^k g_\theta(x)}{\partial\theta^k} \right|_{\theta=0}=a_k x^k\,e^{-1/2},   
\end{equation*}
where
\begin{equation*}
 a_k:=\left(-\frac{1}{2}\right)^k k! \sum _{n=\left\lceil k/2\right\rceil }^k
 \frac1{n!}\,\binom{n}{k-n} (-2)^n. 
\end{equation*}
The sequence $(a_k)$ is the sequence A001464 at http://oeis.org/A001464. 
It appears that $a_k\ne0$ for all $k=0,1,\dots$ except $k=2$. If that is indeed so, then, by $(2)$, we would have 
\begin{equation}
 \int_{\mathbb R}x^kf(x)\,dx=0 \tag{3}
\end{equation}
for all $k=0,1,\dots$ except $k=2$. Letting $y:=x^4$, we see that $x^2=\sqrt y$ can be approximated arbitrarily closely by polynomials in $y=x^4$. So, one would have $(3)$ for all $k=0,1,\dots$, including $k=2$, if $|f(x)|$ decreases fast enough as $|x|\to\infty$. This would imply that $f=0$ almost everywhere. 
A: $g_\theta(x):=e^{-(\theta x-1)^2/2}=\exp(-\frac{1}{2\frac{1}{\theta^2}}[x-\frac{1}{\theta}]^2)$ is the kernel of $N(\frac{1}{\theta},\frac{1}{\theta^2})$ with density $\frac{1}{\sqrt {2\pi \frac{1}{\theta^2}}}\exp\left(-\frac{1}{2\frac{1}{\theta^2}}[x-\frac{1}{\theta}]^2\right)$
which is a curved exponential family. Therefore the guess is that such a family is not complete(See counter example in Example 1 of https://www.stat.tamu.edu/~suhasini/teaching613/chapter2.pdf), the couter-example can be constructed by finding a function $f,h$ such that their Laplace transforms are not the same but $\int_{\mathbb R}g_\theta(x)f(x)\,dx=\int_{\mathbb R}g_\theta(x)h(x)\,dx$. Then the uniqueness theorem of Laplace transform applies. i.e. $f-h\neq0$ yet $\int_{\mathbb R}g_\theta(x)[f-h](x)\,dx=0$
